Math Problem Statement

Past studies have indicated that the percentage of smokers was estimated to be about 30%. Given the new smoking cessation programs that have been implemented, you now believe that the percentage of smokers has reduced. You randomly surveyed 1804 people and found that 507 smoke. Use a 0.05 significance level to test the claim that the percentage of smokers has reduced.

Solution

Here’s a step-by-step breakdown of the problem:


a) Identify the null and alternative hypotheses:

  • Null hypothesis (H₀): The percentage of smokers has not reduced, so the proportion p=0.30p = 0.30.
  • Alternative hypothesis (H₁): The percentage of smokers has reduced, so the proportion p<0.30p < 0.30.

b) Type of hypothesis test:

This is a left-tailed test because the alternative hypothesis is testing if the percentage of smokers is less than 30% (p<0.30p < 0.30).


c) Significance level (α\alpha):

The significance level is α=0.05\alpha = 0.05.


d) Test statistic:

The test statistic for a proportion is calculated as:

z=p^p0p0(1p0)nz = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}}

Where:

  • p^\hat{p} = sample proportion = number of smokerssample size=50718040.2812\frac{\text{number of smokers}}{\text{sample size}} = \frac{507}{1804} \approx 0.2812
  • p0p_0 = hypothesized proportion = 0.30
  • nn = sample size = 1804

Substitute the values:

z=0.28120.300.30×(10.30)1804z = \frac{0.2812 - 0.30}{\sqrt{\frac{0.30 \times (1 - 0.30)}{1804}}}

First, calculate the denominator:

0.30×0.701804=0.211804=0.00011650.0108\sqrt{\frac{0.30 \times 0.70}{1804}} = \sqrt{\frac{0.21}{1804}} = \sqrt{0.0001165} \approx 0.0108

Now calculate zz:

z=0.28120.300.0108=0.01880.01081.74z = \frac{0.2812 - 0.30}{0.0108} = \frac{-0.0188}{0.0108} \approx -1.74

So, the test statistic is approximately z=1.74z = -1.74.


e) P-value:

For a left-tailed test, the p-value is the area to the left of z=1.74z = -1.74 on the standard normal distribution.

Using a Z-table or a calculator, the p-value corresponding to z=1.74z = -1.74 is approximately:

P-value0.0409\text{P-value} \approx 0.0409


f) Decision:

Compare the p-value (0.0409) to the significance level (α=0.05\alpha = 0.05):

  • Since 0.0409<0.050.0409 < 0.05, reject the null hypothesis.

g) Conclusion:

The correct conclusion is:

"There is sufficient evidence to warrant rejection of the claim that the percentage of smokers is less than 30%."


If you’d like, I can provide a detailed explanation for any specific part or help you interpret the results further. Here are some related questions to expand your understanding:

  1. Why is a left-tailed test appropriate for this hypothesis?
  2. What does rejecting the null hypothesis mean in the context of this problem?
  3. How would the results change if the sample size was smaller or larger?
  4. What is the role of the significance level in hypothesis testing?
  5. How would the conclusion differ if the p-value was higher than the significance level?

Tip:

Always check the assumptions of a hypothesis test before calculating! For proportions, ensure np0np_0 and n(1p0)n(1 - p_0) are both greater than 5.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Proportions
Significance Levels
P-Values

Formulas

z = (p̂ - p0) / sqrt((p0 * (1 - p0)) / n)
P-value: Area under the normal curve corresponding to the z-score

Theorems

Central Limit Theorem
Standard Normal Distribution

Suitable Grade Level

Undergraduate Level